Question

Sol. [Key Concept: In a tetrahedral field CFSE $= ( - \frac{3}{5}\ n_{e} + \frac{2}{5}\ n_{t_{2}})\Delta_{t}$; compute for $d^{2},d^{4},d^{6},d^{8}$ and match to the values.]

In tetrahedral complex $\text{CFSE}\text{~} = \left\lbrack - \frac{3}{5}n_{1} + \frac{2}{5}n_{2} \right\rbrack\Delta_{t}$

$$\begin{matrix} & \text{~} \\ & d^{2}:\underset{n_{1} = 2}{\overset{e^{1,1}}{︸}}\underset{n_{2} = 0}{\overset{t^{0,0,0}}{︸}} \\ & \end{matrix}\text{ }\text{~}\text{C.F.S.E}\text{~} = \left\lbrack \frac{- 3}{5}(2) + \frac{2}{5}(2) \right\rbrack\Delta t = - 0.4\Delta t$$

$$d^{4}:\underset{n_{1} = 2}{\overset{e^{1,1}}{︸}}\underset{n_{2} = 2}{\overset{t^{1,1,0}}{︸}}\text{~}\text{ C.F.S.E}\text{~} = \left\lbrack \frac{- 3}{5}(2) \right\rbrack\Delta t = - 1.2\Delta t$$

$$ d^{6}:\underset{n_{1} = 3}{\overset{e^{2,1}}{︸}}\underset{n_{2} = 3}{\overset{t_{2}^{1,1,1}}{︸}}$$; C.F.S.E $= \left\lbrack \frac{- 3}{5}(3) + \frac{2}{5}(3) \right\rbrack\Delta t = - 0.6\Delta t$;

$d^{8}:\underset{n_{1} = 4}{\overset{e^{2,2}}{︸}}\underset{n_{2} = 4}{\overset{t_{2}^{2,1,1}}{︸}}\ \ \ \ \ \ \ \ \ \ $C.F.S.E $= \left\lbrack \frac{- 3}{5}(4) + \frac{2}{5}(4) \right\rbrack\Delta t = - 0.8\Delta t$

A - III, B - IV, C - I, D - II